On Thu, Oct 23, 2008 at 10:29 PM, Monte Milanuk <[EMAIL PROTECTED]> wrote:
> Hello again, and thanks to all of you who extended your help!
> <snip>Maybe I'm not clever enough, but I didn't see how either example
> could be used for doing the addition/subtraction determination. Or rather,
> I found a different way after reading someone else's examples.
>
<snip>
>
# loop thru n times, summing each pass and switching the sign
> # on m each time, and printing x every time through.
> for i in range(n):
> x = x + m*(4.0/(i*2+1))
> m = -m
>
That's probably just as elegant as the way I would have done it; mod
division! :)
In [50]: for x in range(1, 10):
....: print "%d %% 2 = " % (x, ), x % 2
....:
....:
1 % 2 = 1
2 % 2 = 0
3 % 2 = 1
4 % 2 = 0
5 % 2 = 1
6 % 2 = 0
7 % 2 = 1
8 % 2 = 0
9 % 2 = 1
Thus, to adapt it to what you had, declare:
m = -1
And the terms of the loop:
for i in range(1, n+1):
x = x + (m * (i % 2)) * (4.0/(i*2))
At least I'm pretty sure that's syntactically correct. It's 5:47 am and I'm
getting ready for school, so you never know for sure ;)
The great thing about using mod division is that it's /always/ applicable
when you're looking for even/odd status of a number:
In [52]: for x in xrange(-3,3):
....: print "%d %% 2 = " % (x, ), x % 2
....:
....:
-3 % 2 = 1
-2 % 2 = 0
-1 % 2 = 1
0 % 2 = 0
1 % 2 = 1
2 % 2 = 0
Yay for mod division! :)
Glad my advice helped!
-Wayne
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